3 Applications of systems of differential equations

Coupled electrical circuits and mechanical vibrating systems involving several masses in springs offer examples of engineering systems modelled by systems of differential equations.

3.1 Electrical circuits

Consider the RL (resistance/inductance) circuit with a voltage v ( t ) applied as shown in Figure 17.

Figure 17

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If i 1 and i 2 denote the currents in each loop we obtain, using Kirchhoff’s voltage law:

  1. in the right loop: L 1 d i 1 d t + R 2 ( i 1 − i 2 ) + R 1 i 1 = v ( t )
  2. in the left loop: L 2 d i 2 d t + R 2 ( i 2 − i 1 ) = 0
Task!

Suppose, in the above circuit, that

L 1 = 0.8 henry, L 2 = 1 henry, R 1 = 1.4 Ω R 2 = 1 Ω .

Assume zero initial conditions: i 1 ( 0 ) = i 2 ( 0 ) = 0 .

Suppose that the applied voltage is constant: v ( t ) = 100 volts t ≥ 0 .

Solve the problem by Laplace transforms.

Begin by obtaining V ( s ) , the Laplace transform of v ( t ) :

We have, from the definition of the Laplace transform:

V ( s ) = ∫ 0 ∞ 100 e − s t d t = 100 e − s t − s 0 ∞ = 100 s

This is simply the Laplace transform of the step function of height 100.

Now insert the parameter values into the differential equations and obtain the Laplace transform of each equation. Denote by I 1 ( s ) , I 2 ( s ) the Laplace transforms of the unknown currents. (These are equivalent to X ( s ) and Y ( s ) of the theory.):

0.8 d i 1 d t + i 1 − i 2 + 1.4 i 1 = v ( t )

d i 2 d t + i 2 − i 1 = 0

Rearranging and dividing the first equation by 0.8:

d i 1 d t + 3 i 1 − 1.25 i 2 = 1.25 v ( t )

d i 2 d t − i 1 + i 2 = 0

Taking Laplace transforms and inserting the initial conditions i 1 ( 0 ) = 0 , i 2 ( 0 ) = 0 :

( s + 3 ) I 1 ( s ) − 1.25 I 2 ( s ) = 125 s

− I 1 ( s ) + ( s + 1 ) I 2 ( s ) = 0 fractions and finally take the inverse Laplace transform to obtain i 1 ( t ) and i 2 ( t ) :

We find

I 1 ( s ) = 125 ( s + 1 ) s ( s + 1 ∕ 2 ) ( s + 7 ∕ 2 ) = 500 7 s − 125 3 ( s + 1 ∕ 2 ) − 625 21 ( s + 7 ∕ 2 )

in partial fractions.

Hence i 1 ( t ) = 500 7 − 125 3 e − t ∕ 2 − 625 21 e − 7 t ∕ 2

Similarly

I 2 ( s ) = 125 s ( s + 1 ∕ 2 ) ( s + 7 ∕ 2 ) = 500 7 s − 250 3 ( s + 1 ∕ 2 ) + 250 21 ( s + 7 ∕ 2 )

which has inverse Laplace transform:

i 2 ( t ) = 500 7 − 250 3 e − t ∕ 2 + 250 21 e − 7 t ∕ 2

Notice in both cases that i 1 ( t ) and i 2 ( t ) tend to the steady state value 500 7 as t increases.

3.2 Two masses on springs

Consider the vibrating system shown:

Figure 18

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As you can see, the system consists of two equal masses, both m , and 3 springs of the same stiffness k . The governing differential equations can be obtained by applying Newton’s second law (‘force equals mass times acceleration’): (recall that a single spring of stiffness k will experience a force − k y if it is displaced a distance y from its equilibrium.)

In our system therefore

m d 2 y 1 d t 2 = − k y 1 + k ( y 2 − y 1 )

m d 2 y 2 d t 2 = − k ( y 2 − y 1 ) − k y 2

which is a pair of second order differential equations.

Task!

For the above system, if m = 1 , k = 2 and the initial conditions are

y 1 ( 0 ) = 1 y 1 ′ ( 0 ) = 6 y 2 ( 0 ) = 1 y 2 ′ ( 0 ) = − 6

use Laplace transforms to solve the system of differential equations to find y 1 ( t ) and y 2 ( t ) .

Begin by letting Y 1 ( s ) , Y 2 ( s ) be the Laplace transforms of y 1 ( t ) , y 2 ( t ) respectively and take the transforms of the differential equations, inserting the initial conditions:

( s 2 + 4 ) Y 1 − 2 Y 2 = s + 6

− 2 Y 1 + ( s 2 + 4 ) Y 2 = s − 6 partial fractions and finally take inverse Laplace transforms:

Y 1 ( s ) = ( s + 6 ) ( s 2 + 4 ) + 2 ( s − 6 ) ( s 2 + 4 ) 2 − 4 = s s 2 + 2 + 6 s 2 + 6

from which y 1 ( t ) = cos 2 t + sin 6 t

A similar calculation gives y 2 ( t ) = cos 2 t − sin 6 t

We see that the motion of each mass is composed of two harmonic oscillations; the system model was undamped so, on this model, the vibration continues indefinitely.